Answer :13.1 m/s
In this case, the force that enables the car to remain in its circular path is the force of static friction.
$f_s =m \large\frac{v^2}{r}$
The maximum speed the car can have around the curve is the speed at which it is on the verge of skidding outward.
$v_{max} =\sqrt { \large\frac{f_{s, max}r}{m}}$
$\qquad= \sqrt {\large\frac{\mu_s m gr}{m}}$
$\qquad =\sqrt {\mu_s gr}$
$\qquad=\sqrt { (0.500)(9.80\;m/s^2)(35.0)}$
$\qquad= 13.1 \;m/s$